Respuesta :

I think the answer but I'm not totally sure based on what you are asking but I think that the two solutions would be 1 and 0. This is because any number multiplied by 1 would get that number itself and 0 because a number and zero would just get you zero. Hope it helped. 
AL2006
I'm very uncomfortable here.  It's as if there's
some more information lurking just out of sight.

Normally, a "solution" means a value for 'x' that makes
the equation a true statement.

If that's true, and 'c' has to be a whole number, then any
rational value of 'n' gives an infinite number of possible
values for 'x'.

Examples:

-- If  n = 7/3 , then 'c' is a whole number when 'x' is any multiple of 3.
There are an infinite number of those.

-- If n = 1, then 'c' is a whole number when 'x' is any whole number.
There are an infinite number of those.

-- If n = 0, then c=0 whenever 'x' is anything at all.
There are an infinite number of those.

-- Ah hah !  Maybe I found something.
If 'n' is an irrational number, then
     -- c = 1  when  x = 1/n ,  and
     -- c = -1 when  x= -1/n ,
             and those are the only two whole numbers that 'c' can be.

So I  have two conclusions:
#1). I have solved the problem exactly as you stated it.
#2). I can't believe that this is the way it was assigned,
       or that this was the intended answer.
       There's something missing.

Otras preguntas